Equality of ideals for every value of one variable implies they are equal?

Question

Let $J\subseteq I$ be ideals in a polynomial ring $R=\mathbb{F}[x_1, \ldots, x_n, t]$ over a field $\mathbb{F}$ of characteristic zero. Let's write $I_{\lambda}$ for the ideal in $S=\mathbb{F}[x_1, \ldots, x_n]$ obtained by setting $t=\lambda\in \mathbb{F}$ in the ideal $I$ and similarly for $J_{\lambda}$. Suppose $$I_{\lambda}=J_{\lambda} \; \mathrm{ in }\; S \; \mathrm{ for every } \; \lambda \in \mathbb{F}.$$ Does that imply $I=J$ in $R$?

Answer 1

No. Just take $\mathbb{F}=\mathbb{Q}$, $n=0$ and $I=(t),\ J=(t^2)$. Of course, this generalizes to higher values of $n$ trivially.

I think that the radicals of $I$ and $J$ should coincide, though.